# To find the nth Derivative of $9\sqrt{x}$

Can you help me to proof that the nth derivative of $9\sqrt{x}$ is $$(-1)^{(n-1)} \cdot \frac{9(2n-2)!}{(n-1)!} \cdot (4x)^{\frac{1-2n}{2}}$$

I've tried induction but didn't go very far.

Many thanks

• Easiest way will be to find first, second, third and may be 4th derivative and see if you can generalise it. Commented Nov 15, 2015 at 17:25
• Induction works. Take the derivative of your formula and show that it becomes the same formula, but with $n$ replaced by $n+1$. Commented Nov 15, 2015 at 17:28
• I don't thin the four belongs there. Commented Nov 15, 2015 at 17:33

Induction is a very good way. For the induction step,

Say $P(n)$ is true for $n=k$
For $P(k+1)$, $$\frac{d}{dx^{k+1}}(9\sqrt x)$$ $$=\frac{d}{dx}\left(\frac{d}{dx^{k}}(9\sqrt x)\right)$$ $$=\frac{d}{dx} \left[(-1)^{(k-1)} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot (4x)^{\frac{1-2k}{2}}\right]$$ $$=(-1)^{(k-1)} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot \frac{d}{dx} \left[(4x)^{\frac{1-2k}{2}}\right]$$ $$=(-1)^{(k-1)} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot \frac{1-2k}{2} (4x)^{\frac{-1-2k}{2}}\cdot 4$$ $$=(-1)^{[(k+1)-1]} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot (2k-1) (4x)^{\frac{-1-2k}{2}}\cdot 2$$ $$=(-1)^{[(k+1)-1]} \cdot \frac{9(2k-2)!}{(k-1)!} \cdot (2k-1) \cdot \frac{2k}{k} \cdot (4x)^{\frac{-1-2k}{2}}$$ $$=(-1)^{[(k+1)-1]} \cdot \frac{9[2(k+1)-2]!}{[(k+1)-1]!} \cdot (4x)^{\frac{1-2(k+1)}{2}}$$

$P(k+1)$ is true.

• Induction is what makes more sense to me. I got to half of this but you got me through the place where I was stuck. Commented Nov 15, 2015 at 17:42
• You're welcome. Commented Nov 15, 2015 at 17:43
• In your 5th line you are multiplying by 4. How did you do that? If I split the power, as a product I will not get the same. Commented Nov 15, 2015 at 18:27
• I am not multiplying by $4$. I differentiated the entire expression first with respect to $4x$ and then differentiated $4x$ with respect to $x$ to get $4$. Get it? If you split the power as a product, you will get the same. Consider the exponent of $4$. If you split the power as a product, power of $4$ will remain same i.e. $\frac{1-2k}{2}$ and in my working, add the exponents of $4$, you will find it same. Clear?? Commented Nov 16, 2015 at 4:53
• Well I got what you did but still don't understand how splitting it into $4^{\frac{1-2k}{2}} . x^{\frac{1-2k}{2}}$ does not produce the same. Nevertheless, moving on, I do not understand how you go from $(1-2k )$ to $(2k-1)$ and the exponent of $(-1)$ goes from $(k-1)$ to $[(k+1)-1]$. Thanks Commented Nov 16, 2015 at 15:10

Firstly, the 9 is unimportant. Induction is the correct way to proceed, but perhaps we can disguise it in a computation:

\begin{align} \frac{d^n}{dx^n} \left( x^{\frac{1}{2}} \right) &= \frac{d^{n-1}}{dx^{n-1}} \left( \frac{1}{2} x^{\frac{1}{2} -1} \right) \\&= \frac{d^{n-2}}{dx^{n-2}}\left( \frac{1}{2} \left(\frac{1}{2} - 1 \right) x^{\frac{1}{2} -2} \right) \\&= \dots =\frac{d^{n-k}}{dx^{n-k}} \left( \frac{1}{2} \left( \frac{1}{2} -1\right) \left( \frac{1}{2} -2 \right)\dots \left(\frac{1}{2} -(k-1) \right) x^{\frac{1}{2} - k}\right) \\&= \frac{1}{2} \left( \frac{1}{2} - 1 \right) \left( \frac{1}{2} -2 \right)\dots \left(\frac{1}{2} - (n-1) \right) x^{\frac{1}{2} -n }. \end{align} Can you simplify this?

Writing the first four derivatives of $f(x)=x^{1/2}$ reveals

$f^{(1)}(x)=\frac12 x^{-1/2}$

$f^{(2)}(x)=\frac12 \left(-\frac12\right)x^{-3/2}$

$f^{(3)}(x)=\frac12 \left(-\frac12\right)\left(-\frac32\right)x^{-5/2}$

$f^{(4)}(x)=\frac12 \left(-\frac12\right)\left(-\frac32\right)\left(-\frac52\right)x^{-7/2}$

We propose the general term is given by

$$f^{(n)}(x)=(-1)^{n-1}\left(\frac12\right)^n (2n-3)!!x^{-(2n-1)/2}$$

Then, the $n+1$ derivative

\begin{align} f^{(n+1)}(x)&=(-1)^{n-1}\left(\frac12\right)^{n}(-1)\left(\frac12\right) (2n-1)(2n-3)!!x^{-(2n-3)/2}\\\\ &=(-1)^n\left(\frac12\right)^{n+1}(2n-1)!!x^{-(2n-3)/2} \end{align}

which provides proof by induction!

• This is really not a proof...the OP probably did this and deduced the formula. He/she is looking for a proof. Commented Nov 15, 2015 at 17:39
• @rogerl One can easily use induction using the expression of the general term, can one not? And why on earth did you single my answer out as others have posted similarly. Commented Nov 15, 2015 at 17:41
• Sure, but that's what the OP is having trouble with. Commented Nov 15, 2015 at 17:41
• @rogerl I added a bit more to provide the inductive step. Please let me know how I can improve my answer. I really want to give the best answer I can. Commented Nov 15, 2015 at 17:51
• Yes, I think that addresses the OPs question. Commented Nov 15, 2015 at 19:00